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An arrow is shot vertically upward from a platform 21ft high at a rate of 228ft/sec . When will the arrow hit the ground? Use t...Question
An arrow is shot vertically upward from a platform 8ft
high at a rate of 157ft/sec
. When will the arrow hit the ground? Use the formula: h=−16t2+v0t+h0
. (Round your answer to the nearest tenth.)
high at a rate of 157ft/sec
. When will the arrow hit the ground? Use the formula: h=−16t2+v0t+h0
. (Round your answer to the nearest tenth.)
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GPT 3.5
To find when the arrow will hit the ground, we need to find the time when the height is equal to 0.
Using the formula h = -16t^2 + v0t + h0, where h is the height, t is the time, v0 is the initial velocity, and h0 is the initial height, we can substitute the given values:
0 = -16t^2 + 157t + 8
Now, we can solve this quadratic equation to find the value of t. We can either factor it or use the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
In this equation, a is the coefficient of t^2 (-16), b is the coefficient of t (157), and c is the constant term (8).
Plugging in the values, we get:
t = (-157 ± √(157^2 - 4(-16)(8)))/(2(-16))
Simplifying the equation further:
t = (-157 ± √(24649 + 512))/(2(-16))
t ≈ (-157 ± √(25161))/(2(-16))
t ≈ (-157 ± 158.7)/(2(-16))
Now, we can calculate the two solutions:
t ≈ (-157 + 158.7)/(-32)
t ≈ 1.7/(-32) ≈ -0.0531
t ≈ (-157 - 158.7)/(-32)
t ≈ -315.7/(-32) ≈ 9.87
Since time cannot be negative in this context, we discard the negative solution.
Therefore, the arrow will hit the ground approximately 9.9 seconds after it was shot.
Using the formula h = -16t^2 + v0t + h0, where h is the height, t is the time, v0 is the initial velocity, and h0 is the initial height, we can substitute the given values:
0 = -16t^2 + 157t + 8
Now, we can solve this quadratic equation to find the value of t. We can either factor it or use the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
In this equation, a is the coefficient of t^2 (-16), b is the coefficient of t (157), and c is the constant term (8).
Plugging in the values, we get:
t = (-157 ± √(157^2 - 4(-16)(8)))/(2(-16))
Simplifying the equation further:
t = (-157 ± √(24649 + 512))/(2(-16))
t ≈ (-157 ± √(25161))/(2(-16))
t ≈ (-157 ± 158.7)/(2(-16))
Now, we can calculate the two solutions:
t ≈ (-157 + 158.7)/(-32)
t ≈ 1.7/(-32) ≈ -0.0531
t ≈ (-157 - 158.7)/(-32)
t ≈ -315.7/(-32) ≈ 9.87
Since time cannot be negative in this context, we discard the negative solution.
Therefore, the arrow will hit the ground approximately 9.9 seconds after it was shot.
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